
Critical temperature of \[\]\[{H_2}O\] , \[N{H_3}\] , \[C{O_2}\] and \[{O_2}\] are \[647\,K\] , \[405.6\,K\] , \[304.10\,K\] and \[154.2\,K\] respectively. If the cooling starts from \[500\,K\] to their temperature, the gas that liquefies first is:
A. \[{H_2}O\]
B. \[N{H_3}\]
C. \[{O_2}\]
D. \[C{O_2}\]
Answer
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Hint: The critical temperature of a substance can be defined as the highest temperature at which the substance can be converted into liquid. If the temperature is above the critical temperature, the substance in the vapor or gaseous state can no longer be liquefied, regardless of the amount of pressure applied to it. It is denoted by \[{T_c}\].
Complete step by step answer:
We know that below the critical temperature a gas or vapor can be liquefied but above the critical temperature, a gas or vapor cannot be liquefied.
In the question, cooling starts from \[500\,K\] which means we have to decrease the temperature towards the critical temperature. The critical temperature of \[{H_2}O\] is \[647\,K\] which is more than the cooling temperature due to which it can be liquefied.
When cooling starts from \[500\,K\] , the \[N{H_3}\] (ammonia) gas whose critical temperature is \[405.6\,K\] will be liquefies first followed by \[C{O_2}\] and \[{O_2}\] whose critical temperature is \[304.10\,K\] and \[154.2\,K\].
Therefore, the gas that liquefies first is \[N{H_3}\] (ammonia) at \[405.6\,K\].
So, the correct answer is Option B.
Note: It was first observed in 1822 by a French Physicist Charles Cagniard de la Tour that carbon dioxide could be liquified at a temperature of \[{31^ \circ }C\] when 73 atmospheric pressure was applied, but it could not be liquified at higher temperatures, even when pressures above 3000 atmospheric were applied. This maximum temperature at which substances could exist in the liquid phase was later named “Critical Temperature” by Dmitri Mendeleev in the year 1860.
With the concept of critical temperature there comes critical pressure. Critical pressure is the pressure that is applied to liquefy a substance at its critical temperature.
Complete step by step answer:
We know that below the critical temperature a gas or vapor can be liquefied but above the critical temperature, a gas or vapor cannot be liquefied.
In the question, cooling starts from \[500\,K\] which means we have to decrease the temperature towards the critical temperature. The critical temperature of \[{H_2}O\] is \[647\,K\] which is more than the cooling temperature due to which it can be liquefied.
When cooling starts from \[500\,K\] , the \[N{H_3}\] (ammonia) gas whose critical temperature is \[405.6\,K\] will be liquefies first followed by \[C{O_2}\] and \[{O_2}\] whose critical temperature is \[304.10\,K\] and \[154.2\,K\].
Therefore, the gas that liquefies first is \[N{H_3}\] (ammonia) at \[405.6\,K\].
So, the correct answer is Option B.
Note: It was first observed in 1822 by a French Physicist Charles Cagniard de la Tour that carbon dioxide could be liquified at a temperature of \[{31^ \circ }C\] when 73 atmospheric pressure was applied, but it could not be liquified at higher temperatures, even when pressures above 3000 atmospheric were applied. This maximum temperature at which substances could exist in the liquid phase was later named “Critical Temperature” by Dmitri Mendeleev in the year 1860.
With the concept of critical temperature there comes critical pressure. Critical pressure is the pressure that is applied to liquefy a substance at its critical temperature.
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